Diagonal census for the trace-$10$ no-four orbit-matrix case

Let $C$ be a $9\times 9$ matrix over $\mathbb{N}$ whose diagonal entries lie in $\{0,2,4\}$, whose diagonal sums to $10$, and with no diagonal entry equal to $4$. Then exactly five diagonal entries equal $2$ and four equal $0$. Indeed every diagonal entry is $0$ or $2$, so with $c_2$ entries equal to $2$ the trace condition gives $2c_2 = 10$ and the nine diagonal positions give $c_2 + c_0 = 9$. This pins the row-type multiset used by the profile census in the trace-$10$, no-four branch of Wilbrink's Theorem 5.